Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{x - 6}{x + 3} \times \dfrac{x + 9}{x^2 + 3x - 54} $
Solution: First factor the quadratic. $r = \dfrac{x - 6}{x + 3} \times \dfrac{x + 9}{(x - 6)(x + 9)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (x - 6) \times (x + 9) } { (x + 3) \times (x - 6)(x + 9) } $ $r = \dfrac{ (x - 6)(x + 9)}{ (x + 3)(x - 6)(x + 9)} $ Notice that $(x + 9)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ \cancel{(x - 6)}(x + 9)}{ (x + 3)\cancel{(x - 6)}(x + 9)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $r = \dfrac{ \cancel{(x - 6)}\cancel{(x + 9)}}{ (x + 3)\cancel{(x - 6)}\cancel{(x + 9)}} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $r = \dfrac{1}{x + 3} ; \space x \neq 6 ; \space x \neq -9 $